If the equation $a(b-c)x^{2}+b(c-a)x+c(a-b)=0$ has equal roots, where $a+c=15$ and $b=\dfrac{36}{5}$, then $a^{2}+c^{2}$ is equal to:
Step-by-Step Solution
Key Concept: In $a(b-c)x^{2}+b(c-a)x+c(a-b)=0$, the coefficients sum to zero, so $x=1$ is always a root. For equal roots, both roots are $1$, hence the product of roots equals $1$.
Sum of coefficients: $a(b-c)+b(c-a)+c(a-b)=ab-ac+bc-ab+ac-bc=0$. Hence $x=1$ is a root.
Equal roots $\Longrightarrow$ both roots are $1\Longrightarrow$ product of roots $=\dfrac{c(a-b)}{a(b-c)}=1$, i.e. $c(a-b)=a(b-c)$, which simplifies to
$$2ac=b(a+c).$$
With $a+c=15$ and $b=\dfrac{36}{5}$:
$$2ac = \tfrac{36}{5}\cdot 15 = 108\ \Longrightarrow\ ac=54.$$
Therefore
$$a^{2}+c^{2}=(a+c)^{2}-2ac = 225 - 108 = 117.$$
Correct Answer: 117